Misc 15 - Chapter 6 Class 12 Application of Derivatives

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Misc 15 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4𝑟/3 . Given, Radius of sphere = r Let R be the radius of the cone and H be its height. Let ∠ BOC = θ Now, AC = AO + OC H = r + r cos θ And, R = r sin θ H = r (1 + cos θ) We need to maximize volume of cone. Volume of the cone is V = 1/3 𝜋𝑅^2 𝐻 V = 1/3 𝜋𝑟^2 sin^2⁡"θ" r (1 + cos θ) V = 1/3 𝜋𝑟^3 sin^2⁡"θ" (1 + cos θ) Finding 𝑑𝑣/𝑑𝜃 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 [2 sin⁡〖"θ" cos⁡〖"θ" (1+cos⁡〖"θ" )+sin^2⁡〖"θ" (−sin⁡〖"θ" )〗 〗 〗 〗 〗 ] 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 (2 sin⁡〖"θ" cos⁡〖"θ" (1+cos⁡〖"θ" )−sin^3⁡"θ" 〗 〗 〗 ) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 (2 sin⁡〖"θ" cos⁡〖"θ" (1+cos⁡〖"θ" )−sin^3⁡"θ" 〗 〗 〗 ) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (2 cos⁡〖"θ" +〖2 cos^2〗⁡〖"θ" −sin^2⁡"θ" 〗 〗 ) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (2 cos⁡〖"θ" +〖2 cos^2〗⁡〖"θ" −(1−cos^2⁡"θ" 〗 〗)) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (3 cos^2⁡〖"θ" +〖2 𝑐𝑜𝑠〗⁡〖"θ" −1〗 〗 ) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (3 cos^2⁡〖"θ" +〖3 𝑐𝑜𝑠〗⁡𝜃 〗 −cos⁡𝜃 −1) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (3 〖𝑐𝑜𝑠 〗⁡"θ" (cos⁡𝜃+1)−1(cos⁡𝜃+1)) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (3 𝑐𝑜𝑠⁡〖"θ" −1〗 )(𝑐𝑜𝑠⁡〖"θ" +1〗 ) Putting 𝑑𝑣/𝑑𝜃 = 0 1/3 𝜋𝑟^3 sin θ (𝑐𝑜𝑠⁡〖"θ" +1〗 ) (3 𝑐𝑜𝑠⁡〖"θ" −1〗 ) = 0 sin θ (cos θ + 1) (3 cos "θ" − 1) = 0 𝐬𝐢𝐧 θ = 0 θ = 0° θ cannot be 0° for cone 𝐜𝐨𝐬 θ + 1 = 0 cos θ = −1 For cone, 0° < θ < 90° & cos θ is negative in II & III quadrant. So cos θ = −1 is not possible 𝟑 𝐜𝐨𝐬 θ − 1 = 0 cos θ = 1/3 θ = cos−1 1/3 cos θ = 1/3 is possible So, 𝑐𝑜𝑠⁡"θ" = 1/3 Thus, H = r (1 + cos θ) H = r ("1 + " 1/3) H = 4𝑟/3 This value is either maxima or minima. So, finding (𝑑^2 𝑉)/(𝑑𝜃^2 ) 𝑑𝑉/𝑑𝜃 = (1/3 𝜋𝑟^3 sin⁡𝜃 (3 cos2 𝜃 + 2 cos⁡𝜃− 1)) = 1/3 𝜋𝑟^3 [cos⁡𝜃 (3 cos^2⁡𝜃+2 cos⁡𝜃−1)+sin⁡𝜃 (6 cos⁡𝜃 (−𝑠𝑖𝑛)−2 sin⁡𝜃 )] = 1/3 𝜋𝑟^3 [3 cos^3⁡𝜃+2 cos^2⁡𝜃−cos⁡𝜃+sin⁡𝜃 (−6 cos⁡𝜃 sin⁡𝜃−2 sin⁡𝜃 )] = 1/3 𝜋𝑟^3 [3 cos^2⁡𝜃+2 cos^2⁡𝜃−cos⁡𝜃−6 sin^2⁡𝜃 cos⁡𝜃−2 sin^2⁡𝜃 ] Now, cos⁡𝜃=1/3 And sin2 "θ" = 1 − cos2 "θ" = 1 − (1/3)^2= 1 − 1/9 = 8/9 Putting values in (𝑑^2 𝑉)/(𝑑𝜃^2 ) (𝑑^2 𝑉)/(𝑑𝜃^2 ) = 1/3 𝜋𝑟^3 [3(1/3)^3+2(1/3)^2−1/3−6(1/3)(8/9)−2(8/9)] = 1/3 𝜋𝑟^3 [3(1/27)+2(1/9)−1/3−6(8/27)−2(8/9)] = 1/3 𝜋𝑟^3 [1/9+ 2/9−1/3−16/9−16/9] = 1/3 𝜋𝑟^3 [(−32)/9] = (−32𝜋𝑟^3)/27 Thus, (𝑑^2 𝑣)/(𝑑𝜃^2 ) < 0 H = 𝟒𝒓/𝟑 Hence proved